# Natural Number Multiplication Distributes over Addition

## Theorem

The operation of multiplication is distributive over addition on the set of natural numbers $\N$:

$\forall x, y, z \in \N:$
$\left({x + y}\right) \times z = \left({x \times z}\right) + \left({y \times z}\right)$
$z \times \left({x + y}\right) = \left({z \times x}\right) + \left({z \times y}\right)$

## Proof 1

 $\displaystyle \left({x + y}\right) \times z$ $=$ $\displaystyle +^z \left({x + y}\right)$ Definition of multiplication $\displaystyle$ $=$ $\displaystyle \left({+^z x}\right) + \left({+^z y}\right)$ Power of Product of Commuting Elements in Semigroup equals Product of Powers $\displaystyle$ $=$ $\displaystyle x \times z + y \times z$

$\Box$

 $\displaystyle z \times \left({x + y}\right)$ $=$ $\displaystyle +^{x + y} z$ Definition of multiplication $\displaystyle$ $=$ $\displaystyle \left({+^x z}\right) + \left({+^y z}\right)$ Index Laws for Semigroup: Sum of Indices $\displaystyle$ $=$ $\displaystyle z \times x + z \times y$

$\blacksquare$

## Proof 2

We are to show that:

$\left({x + y}\right) \times z = \left({x \times z}\right) + \left({y \times z}\right)$

for all $x, y, z \in \N$.

From the definition of natural number multiplication, we have by definition that:

 $\displaystyle \forall m, n \in \N: \ \$ $\displaystyle m \times 0$ $=$ $\displaystyle 0$ $\displaystyle m \times n^+$ $=$ $\displaystyle \left({m \times n}\right) + m$

Let $x, y \in \N$ be arbitrary.

For all $z \in \N$, let $P \left({z}\right)$ be the proposition:

$\forall x, y \in \N: \left({x + y}\right) \times z = \left({x \times z}\right) + \left({y \times z}\right)$

### Basis for the Induction

$P \left({0}\right)$ is the case:

 $\displaystyle \left({x + y}\right) \times 0$ $=$ $\displaystyle 0$ Definition of Natural Number Multiplication $\displaystyle$ $=$ $\displaystyle 0 + 0$ Definition of Natural Number Addition $\displaystyle$ $=$ $\displaystyle x \times 0 + y \times 0$ Definition of Natural Number Multiplication

and so $P \left({0}\right)$ holds.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k^+}\right)$ is true.

So this is our induction hypothesis:

$\forall x, y \in \N: \left({x + y}\right) \times k = \left({x \times k}\right) + \left({y \times k}\right)$

Then we need to show:

$\forall x, y \in \N: \left({x + y}\right) \times k^+ = \left({x \times k^+}\right) + \left({y \times k^+}\right)$

### Induction Step

This is our induction step:

 $\displaystyle \left({x + y}\right) \times k^+$ $=$ $\displaystyle \left({x + y}\right) \times k + \left({x + y}\right)$ Definition of Natural Number Multiplication $\displaystyle$ $=$ $\displaystyle \left({x \times k}\right) + \left({y \times k}\right) + \left({x + y}\right)$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \left({\left({x \times k}\right) + x}\right) + \left({\left({y \times k}\right) + y}\right)$ Natural Number Addition is Commutative and Associative $\displaystyle$ $=$ $\displaystyle \left({x \times k^+}\right) + \left({y \times k^+}\right)$ Definition of Natural Number Multiplication

So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the Principle of Mathematical Induction:

$\forall x, y, z \in \N: \left({x + y}\right) \times n = \left({x \times z}\right) + \left({y \times z}\right)$

$\Box$

Next we need to show that:

$z \times \left({x + y}\right) = \left({z \times x}\right) + \left({z \times y}\right)$

for all $x, y, z \in \N$.

So:

 $\displaystyle z \times \left({x + y}\right)$ $=$ $\displaystyle \left({x + y}\right) \times z$ Natural Number Multiplication is Commutative $\displaystyle$ $=$ $\displaystyle \left({x \times z}\right) + \left({y \times z}\right)$ from above $\displaystyle$ $=$ $\displaystyle \left({z \times x}\right) + \left({z \times y}\right)$ Natural Number Multiplication is Commutative

Thus we have proved:

$\forall x, y, z \in \N: z \times \left({x + y}\right) = \left({z \times x}\right) + \left({z \times y}\right)$

$\blacksquare$

## Proof 3

In the Axiomatization of 1-Based Natural Numbers, this is rendered:

$\forall x, y, z \in \N_{> 0}:$
$\left({x + y}\right) \times z = \left({x \times z}\right) + \left({y \times z}\right)$
$z \times \left({x + y}\right) = \left({z \times x}\right) + \left({z \times y}\right)$

Using the following axioms:

 $(A)$ $:$ $\displaystyle \exists_1 1 \in \N_{> 0}:$ $\displaystyle a \times 1 = a = 1 \times a$ $(B)$ $:$ $\displaystyle \forall a, b \in \N_{> 0}:$ $\displaystyle a \times \paren {b + 1} = \paren {a \times b} + a$ $(C)$ $:$ $\displaystyle \forall a, b \in \N_{> 0}:$ $\displaystyle a + \paren {b + 1} = \paren {a + b} + 1$ $(D)$ $:$ $\displaystyle \forall a \in \N_{> 0}, a \ne 1:$ $\displaystyle \exists_1 b \in \N_{> 0}: a = b + 1$ $(E)$ $:$ $\displaystyle \forall a, b \in \N_{> 0}:$ $\displaystyle$Exactly one of these three holds: $\displaystyle a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y}$ $(F)$ $:$ $\displaystyle \forall A \subseteq \N_{> 0}:$ $\displaystyle \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0}$

### Left Distributive Law for Natural Numbers

First we show that:

$n \times \left({x + y}\right) = \left({n \times x}\right) + \left({n \times y}\right)$

Let us cast the proposition in the form:

$\forall a, b, n \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\forall a, b \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$

### Basis for the Induction

$\map P 1$ is the case:

 $\displaystyle a \times \paren {b + 1}$ $=$ $\displaystyle \paren {a \times b} + a$ Axiom $B$ $\displaystyle$ $=$ $\displaystyle \paren {a \times b} + \paren {a \times 1}$ Axiom $A$

and so $\map P 1$ holds.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$\forall a, b \in \N_{> 0}: a \times \paren {b + k} = \paren {a \times b} + \paren {a \times k}$

Then we need to show:

$\forall a, b \in \N_{> 0}: a \times \paren {b + \paren {k + 1} } = \paren {a \times b} + \paren {a \times \paren {k + 1} }$

### Induction Step

This is our induction step:

 $\displaystyle a \times \paren {b + \paren {k + 1} }$ $=$ $\displaystyle a \times \paren {\paren {b + k} + 1}$ Axiom $C$ $\displaystyle$ $=$ $\displaystyle \paren {a \times \paren {b + k} } + a$ Axiom $B$ $\displaystyle$ $=$ $\displaystyle \paren {\paren {a \times b} + \paren {a \times k} } + a$ Induction hypothesis $\displaystyle$ $=$ $\displaystyle \paren {a \times b} + \paren {\paren {a \times k} + a}$ Natural Number Addition is Associative $\displaystyle$ $=$ $\displaystyle \paren {a \times b} + \paren {a \times \paren {k + 1} }$ Axiom $B$

The result follows by the Principle of Mathematical Induction.

$\Box$

### Right Distributive Law for Natural Numbers

Then we show that:

$\left({x + y}\right) \times n = \left({x \times n}\right) + \left({y \times n}\right)$

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\forall a, b \in \N_{> 0}: \paren {a + b} \times n = \paren {a \times n} + \paren {b \times n}$

### Basis for the Induction

$\map P 1$ is the case:

 $\displaystyle \paren {a + b} \times 1$ $=$ $\displaystyle a + b$ Axiom $A$ $\displaystyle$ $=$ $\displaystyle \paren {a \times 1} + \paren {b \times 1}$ Axiom $A$

and so $\map P 1$ holds.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$\forall a, b \in \N_{> 0}: \paren {a + b} \times k = \paren {a \times k} + \paren {b \times k}$

Then we need to show:

$\forall a, b \in \N_{> 0}: \paren {a + b} \times \paren {k + 1} = \paren {a \times \paren {k + 1} } + \paren {b \times \paren {k + 1} }$

### Induction Step

This is our induction step:

 $\displaystyle \paren {a + b} \times \paren {k + 1}$ $=$ $\displaystyle \paren {\paren {a + b} \times k} + \paren {a + b}$ Axiom $B$ $\displaystyle$ $=$ $\displaystyle \paren {\paren {a \times k} + \paren {b \times k} } + \paren {a + b}$ Induction hypothesis $\displaystyle$ $=$ $\displaystyle \paren {\paren {a \times k} + a} + \paren {\paren {b \times k} + b}$ Natural Number Addition is Commutative $\displaystyle$ $=$ $\displaystyle \paren {a \times \paren {k + 1} } + \paren {b \times \paren {k + 1} }$ Axiom $B$

The result follows by the Principle of Mathematical Induction.

$\Box$

The result follows.

$\blacksquare$